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77julia77 [94]
11 months ago
13

14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards

Physics
1 answer:
quester [9]11 months ago
8 0

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

brainly.com/question/25239010

#SPJ9

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i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

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v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

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S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

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Putting the value of \frac{dr}{dt}

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[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

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